2 If X Bin 30 045 find i The probability of more than 11 su

2. If X ~ Bin (30, 0.45), find: i) The probability of more than 11 successes ii) The probability of at most 8 failures iii) Give the mean and the standard deviation of X

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

a)
P( X < = 11) = P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1)
= ( 30 11 ) * 0.45^11 * ( 1- 0.45 ) ^19 + .....( 30 0 ) * 0.45^0 * ( 1- 0.45 ) ^30
= 0.2327
P( X > 11) = 1 - P ( X <=11) = 1 -0.2327 = 0.7673

b)
P( X < = 8) = P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 30 8 ) * 0.45^8 * ( 1- 0.45 ) ^22 + ( 30 7 ) * 0.45^7 * ( 1- 0.45 ) ^23 + ( 30 6 ) * 0.45^6 * ( 1- 0.45 ) ^24 + ( 30 5 ) * 0.45^5 * ( 1- 0.45 ) ^25 + ( 30 4 ) * 0.45^4 * ( 1- 0.45 ) ^26 + ( 30 3 ) * 0.45^3 * ( 1- 0.45 ) ^27 + ( 30 2 ) * 0.45^2 * ( 1- 0.45 ) ^28 + ( 30 1 ) * 0.45^1 * ( 1- 0.45 ) ^29 + ( 30 0 ) * 0.45^0 * ( 1- 0.45 ) ^30
= 0.0312

c)
Mean ( np ) =30 * 0.45 = 13.5
Standard Deviation ( npq )= 30*0.45*0.55 = 2.7249